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Working from the bottom up: since k3= k2+ m and k2= k1+ m, k3= k1+ 2m.

 

Then d= c+ k3 so d= c+ k1+ 2m. c= b+ k2 so c= b+ k1+ m and d= b+ 2k1+ 3m.

b= a+ k1 so c= a+ 2k1+ m and d= a+ 3k1+ 3m

 

That is: b= a+ k1, c= a+ 2k1+ m, and d= a+ 3k1+ 3m so we can write everything in terms of a, k1, and m.

 

The real problem is those powers of b, c, and d and the fact that the definition of the rest of the terms is not clear-

After a+ b^2+ c^3+ d^4, I would expect the next term to be a+ b^2+ c^3+ d^4+ e^5 but you have not defined "e"!

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Each part is (a+(n-1)k+1/2(n-2)(n-1)m)^n

 

so the sum is Sum from 1 to i of (a+(n-1)k+1/2(n-2)(n-1)m)^n


Working from the bottom up: since k3= k2+ m and k2= k1+ m, k3= k1+ 2m.

 

Then d= c+ k3 so d= c+ k1+ 2m. c= b+ k2 so c= b+ k1+ m and d= b+ 2k1+ 3m.

b= a+ k1 so c= a+ 2k1+ m and d= a+ 3k1+ 3m

 

That is: b= a+ k1, c= a+ 2k1+ m, and d= a+ 3k1+ 3m so we can write everything in terms of a, k1, and m.

 

The real problem is those powers of b, c, and d and the fact that the definition of the rest of the terms is not clear-

After a+ b^2+ c^3+ d^4, I would expect the next term to be a+ b^2+ c^3+ d^4+ e^5 but you have not defined "e"!

 

I assume the pattern continues e = d + k4 etc and k5=k4+m


[latex]\sum_{n=1}^{i} \left(a+(n-1)\cdot k+\frac{(n-2)(n-1)}{2}\cdot m\right)^n[/latex]

[latex]\sum_{n=1}^{i} \left(a+(n-1)\cdot k+\frac{(n-2)(n-1)}{2}\cdot m\right)^n[/latex]

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Excellent.

Was the problem tough??

 

Five minutes whilst on a phone call - took longer to write the post.

 

- wrote the formula in word

- did multiple find and replace (ie find e replace d+k4; find k3 replace k2+m etc.)

- simplified

- noticed that k coeff went up with n-1

- noticed that m coeff didn't

- did n=6 and n=7 to find out m coeff

- could now tell m coeff was triangular number of n-2

- thought about it till was sure that assumptions looked correct

 

But quite liked the idea - seemed horrifically complicated but resolved down quite quickly; that's assuming I didn't screw up :)

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